A certain basket contains 10 apples 7 of which are red and 3 are green

Let's use some probability rules.

Let's find the probability of selecting a red apple 1st, a red apple 2nd, and a green apple 3rd (aka RRG)
P(red apple 1st AND red apple 2nd AND green apple 3rd) = P(red apple 1st) x P(red apple 2nd) x P(green apple 3rd)
= 7/10 x 6/9 x 3/8
= 7/40

IMPORTANT: selecting a red apple 1st, a red apple 2nd, and a green apple 3rd (aka RRG) is JUST ONE WAY to get 2 red apples and 1 green apple. There's also RGR and GGR.

We already know that P(RRG) = 7/40, which also means P(RGR) = 7/40 and P(GRR) = 7/40
So, P(select 2 red apples and 1 green apple) = P(RRG OR RGR OR GRR)
= P(RRG) + P(RGR) + P(GRR)
= 7/40 + 7/40 + 7/40
= 21/40

Answer: D

Choosing 3 apples from a basket of 10 apples = 10C3
2 red apples and 1 green apple = 7C2 * 3C1
Probability = 7C2 * 3C1 / 10 C3
= 21 * 3 / 120
= 21/40
Option D is the answer

Posted from my mobile device

The number of ways to select 2 red apples is 7C2 = (7 x 6)/(2!) = 21.

The number of ways to select 1 green apple is 3C1 = 3.

So the total number of ways to select 2 red apples and 1 green apple is 21 x 3 = 63.

The total number of ways to select 3 apples from 10 is 10C3 = (10 x 9 x 8)/(3 x 2) = 5 x 3 x 8 = 120.

Thus, the total probability is 63/120 = 21/40.

Answer: D

Total ways = 10C3 = 10*9*8/3*2 = 120
Favorable ways = 7C2*3C1 = 21*3 = 63

Probability = 63/120 = 21/40

IMO D

Probability of occurrence of an event E, P(E), is equal to the total number of favorable occurrences of that event, divided by, the total number of possible occurrences. That is,

P(E) = N(favorable)/N(total).

Calculate the numerator:

N(favorable) = Number of ways of choosing 2 red apples from 7 red apples, multiplied by, the number of ways of choosing 1 green apple from 3 green apples

N(favorable) = 7C2 * 3C1 ...............(1)

Calculate the denominator:

N(total) = Number of ways of choosing 3 apples from 10 apples, of which there are 7 red apples and 3 green apples = (10!)/(7!*3!) ........... (2)

Therefore, the required number of ways of choosing 2 red apples and 1 green apples = [7C2 * 3C1]/[(10!)/(7!*3!) ] = 21/40

ANSWER: D

Note: We are using the concept of combinations here since we are only interesting in choosing a set of elements from a larger set of elements. That is, we are not interested in the ordering of the chosen elements. There are two separate groups of identical elements in each group.

total outcomes = 10C3 = 120

favorable outcomes: 7C2 * 3C1 = 63

63/120 = 21/40

Total Apples: 10

Red: 7 and Green: 3

'3' apples are selected randomly, the total number of ways: \(^{10}{C_3} = 120\)

2 Red [From Red = 7]: \(^7{C_2} = 21\) and 1 Green [From Green = 3] = \(^3{C_1} =3\)

=> Probability: \(\frac{(21 * 3) }{ 120} = \frac{21 }{40}\)

Answer D

Given: A certain basket contains 10 apples, 7 of which are red and 3 are green.
Asked: If 3 different apples are randomly selected, what is the probability that out of those 3 , 2 will be red and 1 will be green ?

Total ways = 10C3 = 120
Favorable ways = 7C2*3C1 = 21*3 = 63
Probability = 63/120 = 21/40

IMO D

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